Determine the molar solubility of Ag2CrO4 in a solution containing 0.153 M AgNO3. The Ksp for Ag2CrO4 is 2.0 × 10^-12.

• A. 7.5 × 10^-6 M
• B. 2.0 × 10^-12 M
• C. 3.9 × 10^-8 M
• D. 1.3 × 10^-10 M

Answer: (D)

Verified step by step guidance

1. Start by writing the balanced dissolution equation for Ag2CrO4: ${\mathrm{Ag}2}_{\mathrm{CrO}4}(s)\to 2\mathrm{Ag}(aq)+\mathrm{CrO}4(aq)^-2$.
2. Express the solubility product constant (Ksp) for Ag2CrO4 in terms of the concentrations of the ions: $K_{\mathrm{sp}}=(2s+0.153)^2\cdot s$, where s is the molar solubility of Ag2CrO4.
3. Since the solution already contains 0.153 M AgNO3, the concentration of Ag⁺ ions from AgNO3 is 0.153 M. Therefore, the total concentration of Ag⁺ ions is $2s+0.153$.
4. Substitute the known Ksp value into the expression: $2.0\times {10}^{-}=(2s+0.153)^2\cdot s$.
5. Solve the equation for s, which represents the molar solubility of Ag2CrO4 in the solution containing 0.153 M AgNO3. This involves algebraic manipulation and potentially using the quadratic formula if necessary.